A taut massless string connects two boxes as shown in the figure. The boxes are placed on an incline plane at θ = 31.00. What is the acceleration of boxes as they move down the incline, given m2 = 1.6 kg, m1 = 7.6 kg, μ2 = 0.10 and μ1 = 0.20.
Please help! I have no idea how to do this. I can sum my forces but i just get an extremely high acceleration.
I will assume ox 1 is uphill so it drags back on box 2 and they accelerate together
Forces down slope
(7.6 +1.6) g sin 31 = 4.74 g
Forces up slope
(7.6*.2 +1.6*.1) g cos 31 = 1.44 g
total mass = 7.6+1.6 = 9.2 kg
so
(4.74-1.44)g = 9.2 a
a = .359 g
if g = 9.81 m/s^2
a = 3.52 m/s^2
Okay, this helped a little! Thank you for answering. This may be a dumb question but where did you get 4.74 and 1.44?
To solve this problem, we can first analyze the forces acting on each box separately.
Let's start by considering box m1. The forces acting on m1 are its weight (mg1) and the tension in the string (T). Since the box is on an inclined plane, we need to consider the component of the weight perpendicular to the plane (mg1 * cos θ) and the component parallel to the plane (mg1 * sin θ). The friction force acting on m1 (f1) opposes its motion down the incline.
Now let's consider box m2. The forces acting on m2 are its weight (mg2) and the tension in the string (T). Since m2 is also on the inclined plane, we need to consider the component of the weight perpendicular to the plane (mg2 * cos θ) and the component parallel to the plane (mg2 * sin θ). The friction force acting on m2 (f2) opposes its motion down the incline.
Now, let's apply Newton's second law to each box separately to find their respective accelerations.
For m1:
Summing the forces in the direction perpendicular to the incline, we have:
mg1 * cos θ - T = 0
Summing the forces in the direction parallel to the incline, we have:
mg1 * sin θ - f1 = m1 * a1
where a1 is the acceleration of m1.
For m2:
Summing the forces in the direction perpendicular to the incline, we have:
mg2 * cos θ - T = 0
Summing the forces in the direction parallel to the incline, we have:
mg2 * sin θ - f2 = m2 * a2
where a2 is the acceleration of m2.
Now, let's find the expressions for the friction forces:
f1 = μ1 * (mg1 * cos θ)
f2 = μ2 * (mg2 * cos θ)
Substituting the expressions for the friction forces into the equations for each box, we have:
mg1 * sin θ - μ1 * (mg1 * cos θ) = m1 * a1
mg2 * sin θ - μ2 * (mg2 * cos θ) = m2 * a2
Now we can solve these equations to find the accelerations a1 and a2. With the given values for m1, m2, μ1, μ2, and θ, we can substitute them into the equations and solve for the accelerations.
Note: It is important to ensure that all quantities are in the appropriate units (e.g. kg for mass, m/s^2 for acceleration, etc.)